hdu 6186 CS Course

CS Course

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 52    Accepted Submission(s): 30

Problem Description

Little A has come to college and majored in Computer and Science.

Today he has learned bit-operations in Algorithm Lessons, and he got a problem as homework.

Here is the problem:

You are giving n non-negative integers a1,a2,⋯,an, and some queries.

A query only contains a positive integer p, which means you 
are asked to answer the result of bit-operations (and, or, xor) of all the integers except ap.

 

 

Input

There are no more than 15 test cases. 

Each test case begins with two positive integers n and p
in a line, indicate the number of positive integers and the number of queries.

2≤n,q≤105

Then n non-negative integers a1,a2,⋯,an follows in a line, 0≤ai≤109 for each i in range[1,n].

After that there are q positive integers p1,p2,⋯,pqin q lines, 1≤pi≤n for each i in range[1,q].

 

 

Output

For each query p, output three non-negative integers indicates the result of bit-operations(and, or, xor) of all non-negative integers except ap in a line.

 

 

Sample Input

3 3 1 1 1 1 2 3

 

 

Sample Output

1 1 0 1 1 0 1 1 0

 

记录前缀后缀跑一遍就可以了。

ac代码：

#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
int a[100001];
int zand[100001];
int zand1[100001];
int zxor[100001];
int zxor1[100001];
int zor[100001];
int zor1[100001];
int main()
{
    int n,q;
    while(~scanf("%d %d",&n,&q))
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            if(i==1)
            {
                zand[i]=a[i];
                zor[i]=a[i];
                zxor[i]=a[i];
            }
            else
            {
                zand[i]=(a[i]&zand[i-1]);
                zor[i]=(a[i]|zor[i-1]);
                zxor[i]=(a[i]^zxor[i-1]);
            }
        }
        zand1[n]=zor1[n]=zxor1[n]=a[n];
        for(int i=n-1;i>=1;i--)
        {
             zand1[i]=(a[i]&zand1[i+1]);
             zor1[i]=(a[i]|zor1[i+1]);
             zxor1[i]=(a[i]^zxor1[i+1]);
        }
        while(q--)
        {
            int x;
            scanf("%d",&x);
            if(x==1)
            {
                cout<<zand1[2]<<' '<<zor1[2]<<' '<<zxor1[2]<<endl;
            }
            else if(x==n)
            {
                cout<<zand[n-1]<<' '<<zor[n-1]<<' '<<zxor[n-1]<<endl;
            }
            else
            {
                cout<<(zand[x-1]&zand1[x+1])<<' '<<(zor[x-1]|zor1[x+1])<<' '<<(zxor[x-1]^zxor1[x+1])<<endl;
            }
        }
    }

    return 0;
}